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# 0.499g 0f CuSO4.xH2O when heated to constant weight gave a residue of 0.346g. th...

### Question

0.499g 0f CuSO4.xH2O when heated to constant weight gave a residue of 0.346g. the value of x is? (Cu = 63.5, S = 32.0, O = 16, H = 1)

### Options

A)
0.5
B)
2.0
C)
3.0
D)
4.0 E)
5.0

### Explanation:

To solve this problem, we need to find the number of water molecules (x) in the hydrated copper sulfate (CuSO4.xH2O) using the given mass information.

First, let's determine the mass of the water that was removed when the hydrated copper sulfate was heated. The initial mass of the hydrated copper sulfate is 0.499 g, and the mass of the residue (anhydrous copper sulfate, CuSO4) after heating is 0.346 g. Therefore, the mass of the water removed is:

0.499 g - 0.346 g = 0.153 g

Now, let's find the number of moles of anhydrous copper sulfate (CuSO4) and water (H2O) based on their masses and molar masses:

Moles of CuSO4 = mass / molar mass = 0.346 g / (63.5 g/mol + 32.0 g/mol + 4 × 16 g/mol) = 0.346 g / 159.5 g/mol ≈ 0.00217 mol

Moles of H2O = mass / molar mass = 0.153 g / (2 × 1 g/mol + 16 g/mol) = 0.153 g / 18 g/mol ≈ 0.00850 mol

Now we can find the ratio of moles of water to moles of anhydrous copper sulfate:

Ratio = moles of H2O / moles of CuSO4 = 0.00850 mol / 0.00217 mol ≈ 3.92

Since the ratio is approximately 4, the value of x is 4, which means there are 4 water molecules per molecule of anhydrous copper sulfate in the hydrated compound CuSO4.xH2O. Therefore, the correct answer is Option D: 4.0.